Relativna atomska masa
Submitted by Biologist on Thu, 02/13/2014 - 13:39

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Elementi A i B grade sa O2 dvije veze ABO3 i AB2O5. AB03 sadrzi 17,36%O2 i AB02O5 sadrzi 20,01% O2.
Izracunati relativnu atomsku masu od A i B??
Ar(A)=91,312 ...Zr i od B=137,161 ..Ba
w=Ar/Mr 17,36= 3*15,999
w=Ar/Mr 17,36= 3*15,999/Mr Mr= 276,48 20,01=5*15,999/Mr Mr= 399,775
w0+WA+WB=100% Mr= A+B+3*15,999 Mr= A+B+ 5*15,999
WA+WB= 100%-17,36% 276,48= A+B+ 47,997 399,775= A+2B+ 79,955
WA+WB = 82,64% A+B= 228,483 A+2B= 319,78
A= 228,483- B
228,483- B+ 2B= 319,78 A= 228,483-91,297
B= 91,297 A = 137,186
Ej ljudi, ja sam uradila ovako, ali zaminjeni mi A i B...
I meni ispada isti rezultat
I meni ispada isti rezultat kao kod tebe .Da nije greska u resenju ili si ti permutovala molarne mase A i B?